Let \begin{equation} \omega:\mathbb{R}^n\rightarrow(\mathbb{R}^n)^* \end{equation} be a continuous differential form \begin{equation} \omega(\boldsymbol{x})=\sum_{i=1}^n a_i(\boldsymbol{x})dx_i \quad \forall \boldsymbol{x}\in\mathbb{R}^n \end{equation} and:
\begin{equation} \psi:[a,b]\rightarrow\mathbb{R}^n \end{equation} a regular curve.
We can define the line integral of $\omega$ over $\psi$:
\begin{equation} \int_{\psi}\omega=\int_a^b \sum_{i=1}^{n}a_i(\psi(t))\psi'_i(t)dt \end{equation}
Now, let \begin{equation} \boldsymbol{X}=F(\boldsymbol{x}) \end{equation} be an invertible coordinate transformation.
Let $\Gamma=F(\psi)$ be the image curve and consider $\tilde{\omega}(\boldsymbol{X})=\omega(F^{-1}(\boldsymbol{X}))$ (thus $\omega$ as a function of the new coordinates). Is it true that
\begin{equation} \int_{\Gamma} \tilde{\omega}=\int_{\psi}\omega \end{equation}
? If it is true, how do you prove it? Is there a theorem for it?